#### Answer

$$y=e^{ 2x} $$

#### Work Step by Step

This is a linear equation and has the integrating factor as follows
$$\alpha(x)= e^{\int P(x)dx}=e^{\int (-1)dx}=e^{-x}$$
Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int \alpha(x)Q(x)dx +C\right)\\ &=e^x\left( \int e^{x}dx +C\right)\\ &=e^x\left( e^{x} +C\right) \end{align}
At $x=0$, $y=1$, and we have $C=0$
Thus, the general solution is
$$y=e^{ 2x} $$